# Limit involving probability

Let $\mu$ be any probability measure on the interval $]0,\infty[$. I think the following limit holds, but I don't manage to prove it: $$\frac{1}{\alpha}\log\biggl(\int_0^\infty\! x^\alpha d\mu(x)\biggr) \ \xrightarrow[\alpha\to 0+]{}\ \int_0^\infty\! \log x\ d\mu(x)$$ In probabilistic terms it can be rewritten as: $$\frac{1}{\alpha}\log\mathbb{E}[X^\alpha] \ \xrightarrow[\alpha\to 0+]{}\ \mathbb{E}[\log X]$$ for any positive random variable $X$.

Can you help me to prove it?

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We assume that there is $\alpha_0>0$ such that $\int_0^{+\infty}x^{\alpha_0} d\mu(x)$ is finite. Let $I(\alpha):=\frac 1{\alpha}\log\left(\int_0^{+\infty}x^\alpha d\mu(x)\right)$ and $I:=\int_0^{+\infty}\log xd\mu(x)$.
Since the function $t\mapsto \log t$ is concave, we have $I(\alpha)\geqslant I$ for all $\alpha$.
Now, use the inequality $\log(1+t)\leqslant t$ and the dominated convergence theorem to show that $\lim_{\alpha\to 0^+}\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)=I$. Call $J(\alpha):=\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)$. Then $$I\leqslant I(\alpha)\leqslant J(\alpha).$$
I agree with your statements. But then how can I use then to prove that $I(\alpha)\xrightarrow[\alpha\to 0+]{}I$? – qwertyuio Feb 27 '13 at 11:13
And the inequality $I(\alpha)\leq J(\alpha)$ holds since in general $\log y\leq y-1$. Ok, now it's clear: thank you very much! – qwertyuio Feb 27 '13 at 11:44